# Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$

I would like to try and prove

$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$

using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.

I found a quick way by doing this:

$$LHS = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\cos x} \cdot \frac{1 - \tan x + \sec x}{1 - \tan x + \sec x}= \frac{1+\sin x+\cos x}{1-\sin x+\cos x} = RHS$$

but I feel that this is not a good way because I am manipulating the denominator of the LHS somewhat artificially, because I know it must be, in the end, $1-\sin x+\cos x$.

Does anyone have a better way of doing this?

$\frac{1+\sin x}{\cos x}$

$=\frac{(1+\sin x)(1-\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$

$=\frac{1-\sin^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$

$=\frac{\cos^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$

$=\frac{(\cos x)(1+\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$

$=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$

Using $\cos^2x+\sin^2x=1$, we have $$(1+\sin x)(1-\sin x+\cos x)=1-\sin^2x+(1+\sin x)\cos x=\cos x(\cos x+1+\sin x).$$ Now divide the LHS by $1-\sin x+\cos x$ and the RHS by $\cos x$ to get the result.

For $b\not=0$ we have: \begin{align*}\frac{1+a}{b} &= \frac{1+a+b}{1-a+b}\qquad&\iff \\ (1+a)(1-a+b) &= b(1+a+b)\qquad&\iff\\ 1-a^2+b(a+1) &= b^2 + b(a+1)\qquad&\iff\\ a^2+b^2&=1 \end{align*} So your equation is an alternate way to characterize $\cos^2 x+\sin^2 x = 1$

Let's call $$a=\dfrac{1+s}c$$

Replacing $$1$$ by $$c^2+s^2$$ gives $$\ a=\dfrac{c^2+s^2+s}c=c+s\dfrac{1+s}c=c+sa\iff a(1-s)=c$$

So we have also $$a=\dfrac c{1-s}$$

Now we can use the identity $$\frac pq=\frac rs=x\implies \frac{p+r}{q+s}=\frac{qx+sx}{q+s}=x$$

Thus $$a=\dfrac{(1+s)+c}{c+(1-s)}$$